Question

Evaluate : $\int \frac{x^3}{(x-1)(x^2+1)}dx.$

OR

Evaluate $\int \frac{sinx-xcosx}{x(x+sinx)}$ dx.

Answer 1

$\begin{array}{cc}\text{Let I}=\int \frac{x^3}{(x-1)(x^2+1)}dx& \Rightarrow I\int (1+\frac{x^2-x+1}{(x-1)(x^2-1)})dx=\\ \Rightarrow I=\int 1dx+\int (\frac{x^2+1}{(x-1)(x^2-1)}-\frac{x}{(x-1)(x^2-1)})dx& \Rightarrow I=x+\int \frac{1}{x-1}dx-\int \frac{x}{(x-1)(x^2-1)}dx\\ \Rightarrow I=x+log\left|\begin{array}{c}x-1\end{array}\right|-I_1\dots \left(A\right)& \text{Now}{I}_1=\int \frac{x}{(x-1)(x^2+1)}dx\\ \text{Consider}\frac{x}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{2Bx}{x^2+1}+\frac{C}{x^2+1}& \Rightarrow x=A(x^2+1)+2Bx(x-1)+C(x-1)\end{array}$

On equating the coefficients of like terms on both the sides, we get : $A=\frac{1}{2},B=-\frac{1}{4},C=\frac{1}{2}$

$\therefore I=x+log\left|\begin{array}{c}x-1\end{array}\right|-(\frac{1}{2}\int \frac{dx}{x-1}-\frac{1}{4}\int \frac{2xdx}{x^2+1}+\frac{1}{2}\int \frac{dx}{x^2+1})$

$\Rightarrow I=x+log\left|\begin{array}{c}x-1\end{array}\right|-\frac{1}{2}log\left|\begin{array}{c}x-1\end{array}\right|+\frac{1}{4}log\left|\begin{array}{c}{x}^2+1\end{array}\right|-\frac{1}{2}ta{n}^{-1}x+C$

Therefore, $I=x+\frac{1}{2}log\left|\begin{array}{c}x-1\end{array}\right|+\frac{1}{4}-log\left|\begin{array}{c}{x}^2-1\end{array}\right|-\frac{1}{2}ta{n}^{-1}x+C.$

OR

Let I = $\int \frac{sinx-xcosx}{x(x+sinx)}-\frac{x+xcosx}{x(x+sinx)}dx\Rightarrow I=\int \frac{x+sinx-x-xcosx}{x(x+sinx)}dx$

$\Rightarrow I=\int \left(\frac{x+sinx}{x(x+sinx)}\right)-\frac{x+xcosx}{x(x+sinx)}dx\Rightarrow I=\int \frac{1}{x}dx-\int \frac{1+cosx}{x+sinx}dx$

Put $x+sinx=t\Rightarrow (1+cosx)dx=\text{dt in 2nd integral}\therefore I=log\left|\begin{array}{c}x\end{array}\right|-\int \frac{dt}{t}$

$\Rightarrow I=log\left|\begin{array}{c}x\end{array}\right|-log\left|\begin{array}{c}t\end{array}\right|+C\therefore I=log\left|\begin{array}{c}x\end{array}\right|-log\left|\begin{array}{c}x+sinx\end{array}\right|+C.$